\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 186, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/186\hfil
spectrum analysis of a wave equation]
{Effect of Kelvin-Voigt damping on spectrum analysis of a wave equation}

\author[L. Lu, L. Zhao, J. Hu \hfil EJDE-2018/186\hfilneg]
{Liqing Lu, Liyan Zhao, Jing Hu}

\address{Liqing Lu (corresponding author) \newline
School of mathematical sciences, 
Shanxi University, Taiyuan 030006, China}
\email{lulq@sxu.edu.cn}

\address{Liyan Zhao \newline
School of mathematical sciences, 
Shanxi University, Taiyuan 030006, China}
\email{494531816@qq.com}

\address{Jing Hu  \newline
School of mathematical sciences, 
Shanxi University, Taiyuan 030006, China}
\email{1556357154@qq.com}

\dedicatory{Communicated by Goong Chen}

\thanks{Submitted September 9, 2018. Published November 19, 2018.}
\subjclass[2010]{35L05, 93C20, 35B35}
\keywords{Wave equation; Riesz basis; spectrum-determined growth condition;
\hfill\break\indent  Kelvin-Voigt damping} 

\begin{abstract}
 This article concerns a one-dimensional wave equation with a
 small amount of Kelvin-Voigt damping. We give a detailed spectrum analysis
 of the system operator, from which we show that the generalized eigenfunction
 forms a Riesz basis for the state Hilbert space. That is, the precise and
 explicit expression of the eigenvalues is deduced and the spectrum-determined
 growth condition is established. Hence the exponential stability of the system
 is obtained.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section] 
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks


\section{Introduction and statement of main results}

In the past few decades, stimulated by
a large quantity of applications of smart materials,
there has been an increasing
research  on elastic system with viscoelastic damping.
When the smart materials are added into
the elastic structures, the exponential stability of
the elastic system has attracted many research interests,
especially for a linear vibration system governed by
one-dimensional wave and Euler-Bernoulli beam
equations. The results of exponential stability
by bounded viscous damping can be found in \cite{b1,c1,f1,l1}.
As a kind of unbounded viscoelastic damping
and internal material damping,
Kelvin-Voigt damping presents in all realistic materials
and is specially important.
The exponential stability of one-dimensional wave and Euler-Bernoulli beam
equations with Kelvin-Voigt damping is discussed in \cite{g2,k1,l2,l3}
and \cite{a1} for multi-dimensional case.
The research of spectral analysis and Riesz basis can
be found in \cite{g1,g3,g4,r1}.

In infinite-dimensional control theory,
Riesz basis property is one of the most wanted properties,
particular for elastic vibrating systems for
which the Riesz basis property is significant both theoretically
and practically.
Usually, the Riesz basis property will lead to
the establishment such as the spectrum determined
growth condition, and the exponential stability of the system.

In this work we consider the  wave equation with a small
amount of Keivin-Voigt damping,
\begin{equation}
\begin{gathered}
w_{tt}(x,t)=(1+d\partial_t)(w_{xx}(x,t)-cw(x,t)), \quad 0<x<1,\; t>0,\\
w_x(0,t)=0,\quad w(1,t)=0,\quad t>0,
\end{gathered}\label{e1.1}
\end{equation}
where $c>0$ is a system parameter and $d>0$ is
a small Kelvin-Voigt damping coefficient.

The energy function for\eqref{e1.1} is
\begin{equation}
E(t)=\frac{1}{2}\int_0^1[w^2_t(x,t)+w^2_x(x,t)+cw^2(x,t)]dx.\label{e1.2}
\end{equation}
Formally, it is found that
\begin{equation}
\frac{d}{dt}E(t)=-d\int_0^1w_{xt}^2(x,t)dx-cd\int_0^1w_{t}^2(x,t)dx\leq0.
\label{e1.3}
\end{equation}
So $E(t)$ is non-increasing.

Let $H_R^1(0,1)=\{f(x)\in H^1(0,1)|f(1)=0\}$.
We consider system \eqref{e1.1} in the energy state space
$\mathcal{H}=H_R^1(0,1)\times L^2(0,1)$ with the inner product induced norm:
\begin{equation}
\|(f,g)\|^2=\int_0^1[|f'(x)|^2+|g(x)|^2+c|f(x)|^2]dx.\label{e1.4}
\end{equation}
Then\eqref{e1.1} can be written as an evolutionary equation in
$\mathcal{H}$,
\begin{equation}
\frac{d}{dt}Y(t)=\mathcal{A}Y(t),\label{e1.5}
\end{equation}
where $Y(t)=(w,w_t)$ and the operator $\mathcal{A}$ is given by
\begin{equation}
\begin{gathered}
\mathcal{A}(f(x),g(x))=(g(x),f''(x)-cf(x)+dg''(x)-cdg(x)),\\
\mathcal{D}(\mathcal{A})=\{(f,g)\in[H^2(0,1)\cap H_R^1(0,1)]^2: f'(0)=0,\; g'(0)=0\}.
\end{gathered}\label{e1.6}
\end{equation}
The following Lemma is straightforward.

\begin{lemma} \label{lem1.1}
Let $\mathcal{A}$ be given by \eqref{e1.6}.
Then its adjoint $\mathcal{A}^*$ has the  form
\begin{equation}
\begin{gathered}
\mathcal{A}^*(f(x),g(x))=(-g(x),-f''(x)+cf(x)+dg''(x)-cdg(x)),\\
\mathcal{D}(\mathcal{A}^*)=\{(f,g)\in[H^2(0,1)\cap H_R^1(0,1)]^2:
f'(0)=0,\; g'(0)=0\}.
\end{gathered}\label{e1.7}
\end{equation}
\end{lemma}

The following Definitions are brought from \cite[Chapter 2]{c2}.

\begin{definition} \label{def1.1}\rm
 A sequence of vectors $\{\phi_n,\; n\geq1\}$ in a Hilbert space
$\mathcal{H}$ forms a Riesz basis for $\mathcal{H}$ if the
following two conditions hold:
\begin{itemize}
\item[(i)] $\overline{\rm span}_{n\geq1}\{\phi_n\}=\mathcal{H}$;

\item[(ii)] There exist positive constants $m$ and $M$ such that
for arbitrary $N \in\mathbb{N}$ and arbitrary
scalars $\alpha_n$, $n=1,\ldots,N$, such that
$$
m\sum_{n=1}^{N}|\alpha_n|^2\leq\|\sum_{n=1}^{N}\alpha_n\phi_n\|^2
\leq M\sum_{n=1}^{N}|\alpha_n|^2.
$$
\end{itemize}
\end{definition}

\begin{definition} \label{def1.2} \rm
Suppose that $\mathcal{A}$ is a linear closed operator on a Hilbert space,
$\mathcal{H}$, with simple eigenvalues $\{\lambda_n,\ n\geq1\}$ form a
Riesz basis in $\mathcal{H}$. If the closure of $\{\lambda_n,\ n\geq1\}$
is totally disconnected, then we call $\mathcal{A}$ a Riesz-spectral
operator.
\end{definition}


\begin{remark} \label{rmk1.1} \rm
One can define a Riesz basis for $\mathcal{H}$
comprised of a sequence of vectors $\phi_n$ belonging to a countable
subset of $\mathbb{Z}=\{0,\pm1,\pm2,\ldots\}$.
\end{remark}

The following two theorems will be proved in the next section.

\begin{theorem} \label{thm1.1}
Let $\mathcal{A}$ and $\mathcal{A}^*$ be given by\eqref{e1.6} and\eqref{e1.7}
respectively. Then $\mathcal{A}$ and $\mathcal{A}^*$ are
dissipative, and hence $\mathcal{A}$ generates a $C_0$-semigroup of contractions
 $e^{\mathcal{A}t}$ on  $\mathcal{H}$.
 Moreover, $\mathcal{A}^{-1}$ exists.
\end{theorem}

\begin{theorem} \label{thm1.2}
Let $\mathcal{A}$ be defined by\eqref{e1.6}. Then
there is a sequence of generalized eigenvectors of $\mathcal{A}$ which
forms a Riesz basis for state space $\mathcal{H}$. Hence
semigroup $e^{\mathcal{A}t}$ generated
by $\mathcal{A}$ is exponentially stable if $d^2c<4$.
\end{theorem}

\subsection{Proof of main results}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Given any $(f,g)\in\mathcal{D}(\mathcal{A})$. We have
\begin{align*}
\langle\mathcal{A}(f,g),(f,g)\rangle
&= \langle(g,f''-cf+dg''-cdg),(f,g)\rangle\\
&=\int_0^1 [f''(x)\overline{g(x)}
-\overline{f''(x)}g(x)]dx+c\int_0^1[\overline{f(x)}g(x)
-f(x)\overline{g(x)}]dx\\
&\quad -d\int_0^1[g'(x)]^2dx-cd\int_0^1[g(x)]^2dx,
\end{align*}
and hence
\begin{equation}
\mathcal{R}e\langle\mathcal{A}(f,g),(f,g)\rangle
=-d\int_0^1[g'(x)]^2dx-cd\int_0^1[g(x)]^2dx\leq0.\label{e2.1}
\end{equation}
Similarly for any $(u,v)\in\mathcal{D}(\mathcal{A}^*)$,
\begin{align*}
&\langle\mathcal{A}^*(u,v),(u,v)\rangle \\
&=\langle(-v,-u''+cu+dv''-cdv),(u,v)\rangle\\
&=\int_0^1 [\overline{u''(x)}v(x)
-u(x)\overline{v''(x)}]dx+c\int_0^1[u(x)\overline{v(x)}
-\overline{u(x)}v(x)]dx\\
&\quad -d\int_0^1[v'(x)]^2dx-cd\int_0^1[v(x)]^2dx,
\end{align*}
and hence
\begin{equation}
\mathcal{R}e\langle\mathcal{A}^*(u,v),(u,v)\rangle
=-d\int_0^1[v'(x)]^2dx-cd\int_0^1[v(x)]^2dx\leq0.\label{e2.2}
\end{equation}
Therefore, both $\mathcal{A}$ and $\mathcal{A}^*$
are dissipative. By the Lumer-Phillips Theorem \cite{p1},
$\mathcal{A}$ generates a $C_0$-semigroup
$e^{\mathcal{A}t}$ of contractions in $\mathcal{H}$.

Now we show that $\mathcal{A}^{-1}$ exists.
For any given $(f_1,g_1)\in\mathcal{H}$, solving
$$
\mathcal{A}(f,g)=(g,f''-cf+dg''-cdg)=(f_1,g_1)
$$
gives $g(x)=f_1(x)$ with $f$ satisfying
\begin{gather*}
f''(x)-cf(x)=g_1(x)+cdf_1(x)-df_1''(x) ,\\
f(1)=0,\quad f'(0)=0.
\end{gather*}
The solution of the above ODE has the form
$$
f(x)=\int_0^1[g_1(s)+cdf_1(s)-df_1''(s)]k(x,s)ds,
$$
where
\begin{equation}
k(x,s)=\frac{1}{\sqrt{c}(1+e^{-2\sqrt{c}})}
\begin{cases}
(e^{\sqrt{c}(s-2)}-e^{-\sqrt{c}s})\cosh(\sqrt{c}x), &x<s,\\
\cosh(\sqrt{c}s)(e^{\sqrt{c}(x-2)}-e^{-\sqrt{c}x}), &x>s.
\end{cases} \label{e2.3}
\end{equation}
From this and $g(x)=f_1(x)$, we get the unique
$(f,g)\in\mathcal{D}(\mathcal{A})$. Hence, $\mathcal{A}^{-1}$ exists.
This completes the proof.
\end{proof}

Let us now consider the eigenvalue problem of $\mathcal{A}$.
It is seen that $\mathcal{A}(f,g)=\lambda(f,g)$, where
$(f,g)\in\mathcal{D}(\mathcal{A}),\ \lambda\in\mathbb{C}$, if
and only if $g(x)=\lambda f(x)$, and $f(x)\neq0$ satisfies the
 eigenvalue problem
\begin{equation}
\begin{gathered}(1+d\lambda)f''(x)-
(\lambda^2+cd\lambda+c)f(x)=0,\\
f(1)=0,\quad f'(0)=0.
\end{gathered}\label{e2.4}
\end{equation}

To prove Theorem \ref{thm1.2} we  first
prove the following lemma, which gives the precise and explicit
expression of the eigenvalues of the system operator $\mathcal{A}$.

\begin{lemma} \label{lem2.1}
Let $\mathcal{A}$ be defined by\eqref{e1.6}. Then
 $\mathcal{R}e\lambda<0$ for every $\lambda\in\sigma_p(\mathcal{A})$.
Furthermore,
\smallskip

\noindent\textrm{(i)} the point spectrum $\sigma_p(\mathcal{A})$ has three
families:
\begin{align*}
\sigma_p(\mathcal{A})
&=\big\{\lambda_{1n},\ n\geq N+1,\ n\in\mathbb{Z}\big\}
 \cup \big\{\lambda_{2n},\ n\geq N+1,\ n\in\mathbb{Z}\big\}\\
&\quad \cup\big\{\lambda_{3n},\ \bar{\lambda}_{3n},\
0\leq n\leq N-1,\ n\in\mathbb{Z}\big\}
 \cup\big\{\lambda_N\big\},
\end{align*}
where
\begin{equation}
\begin{gathered}
\lambda_{1n}=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{\Delta_n}], \quad
 n\in\mathbb{Z},\; n\geq N+1,\\
\lambda_{2n}=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{\Delta_n}], \quad
 n\in\mathbb{Z},\; n\geq N+1,\\
\lambda_{3n}= \frac{1}{2}[d(\sigma_n^2-c)+\sqrt{-\Delta_n}i],\quad
 n\in\mathbb{Z},\; 0\leq n\leq N-1,\\
\lambda_N=-\frac{2}{d},
\end{gathered}\label{e2.5}
\end{equation}
where the algebraic multiplicity of $\lambda_N$ is two, the algebraic multiplicity
of other eigenvalues is one. The quantities $\sigma_n$, $\Delta_n$, $N$ will be
given later.

Moreover there are two sets of the eigenvalues of the system operator
$\mathcal{A}$:
For lower $n$ the eigenvalues reside on the circle
\begin{equation}
[\mathcal{R}e(\lambda_n)+\frac{1}{d}]^2+[\mathcal{I}m(\lambda_n)]^2
=\frac{1}{d^2},\label{e2.6}
\end{equation}
and for higher $n$ the eigenvalues are real, with one
branch accumulating towards $-\frac{1}{d}$ as $n\to\infty$
and the other branch converging to $-\infty$.
\smallskip

\noindent \textrm{(ii)} For any $\lambda_n\in\sigma_p(\mathcal{A})$, $n\neq N$,
there is only one associated (linearly independent)
eigenfunction which takes the form $(\frac{1}{\lambda_n}f_n,f_n)$, where
$f_n$ is found to be
\begin{equation}
f_n(x)=\cos[(n+\frac{1}{2})\pi x],\quad n=0,1,2,\dots.\label{e2.7}
\end{equation}
\end{lemma}

\begin{remark} \label{rmk2.1} \rm
It should be noted that a wave equation
with Kelvin-Voigt damping cannot be categorized
as a hyperbolic PDE. With at most a finite number of
conjugate-complex eigenvalues in its spectrum, such a PDE
should be categorized as a parabolic/hyperbolic hybrid.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lem2.1}]
Denote $\sigma^2=\frac{\lambda^2+cd\lambda+c}{1+d\lambda}$.
Then, by the equivalence, the  ODE
\begin{equation}
\begin{gathered}
f''(x)-\sigma^2f(x)=0,,\\
f(1)=f'(0)=0,
\end{gathered}\label{e2.8}
\end{equation}
has a non-trivial solution
if and only if
$$
\sigma_n=(n+\frac{1}{2})\pi i,\quad n=0,\pm1,\pm2,\dots.
$$
The corresponding eigenfunctions are
$$
f_n(x)=\cos[(n+\frac{1}{2})\pi x],\quad n=0,\pm1,\pm2,\dots.
$$
Hence the eigenvalues of $\mathcal{A}$ satisfy the characteristic equation
\begin{equation}
\lambda_n^2+d(c-\sigma_n^2)\lambda_n+c-\sigma_n^2=0.\label{e2.9}
\end{equation}
Setting $\Delta_n=d^2(c-\sigma_n^2)^2-4(c-\sigma_n^2)$, we have
$$
\Delta_n=d^2(c-\sigma_n^2)[(n+\frac{1}{2})^2\pi^2-(\frac{4}{d^2}-c)].
$$
Considering  $d>0$ as a small Kelvin-Voigt damping coefficient such that
$\frac{4}{d^2}>c$, there exists a positive integer $N$ such that
\begin{equation}
(N+\frac{1}{2})^2\pi^2=\frac{4}{d^2}-c<(N+\frac{3}{2})^2\pi^2.\label{e2.10}
\end{equation}
Then for $n\in\mathbb{Z},\ n\geq N+1$ or $n\leq-N-2$, we have
$\Delta_n>0$, and in this case, the eigenvalues of $\mathcal{A}$ are
\begin{equation}
\begin{gathered}
\lambda_{1n} =\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{\Delta_n}],\\
\lambda_{2n} =\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{\Delta_n}],
\end{gathered}\label{e2.11}
\end{equation}
where
\begin{equation}
\begin{gathered}
\sigma_n=(n+\frac{1}{2})\pi i,\\
\Delta_n=d^2(c-\sigma_n^2)^2-4(c-\sigma_n^2).
\end{gathered} \label{e2.12}
\end{equation}
Noticing that $\sigma^2_n=\sigma^2_{-n-1}$, and
$\lambda_{n}=\lambda_{(-n-1)}$,   for any
 $n\in\mathbb{Z}$, $n\geq N+1$, we have that
\begin{equation}
\begin{gathered}
\lambda_{1n}
=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{\Delta_n}],\\
\lambda_{2n}
=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{\Delta_n}].
\end{gathered}\label{e2.13}
\end{equation}
It is easy to check that $\mathcal{R}e\lambda_{1n}<0$, and
 $\mathcal{R}e\lambda_{2n}<0$.
Furthermore, for any $n\in\mathbb{Z}$ with $n\geq N+1$,
\begin{align*}
\lambda_{1n}
&=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{\Delta_n}]\\
&=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{d^2(c-\sigma_n^2)^2-4(c-\sigma_n^2)}]\\
&=\frac{d}{2}(\sigma_n^2-c)[1-\sqrt{1-\frac{4}{d^2(c-\sigma_n^2)}}],\\
\lambda_{2n}
&=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{\Delta_n}]\\
&=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{d^2(c-\sigma_n^2)^2-4(c-\sigma_n^2)}],
\end{align*}
where $c-\sigma_n^2=c+(n+\frac{1}{2})^2\pi^2$  converges to
$+\infty$ as $n\to\infty$. Moreover
$$
d(\sigma_n^2-c)<\lambda_{2n}<\frac{1}{2}d(\sigma_n^2-c).
$$
Then $\lambda_{1n}$ accumulates towards $-1/d$ and
$\lambda_{2n}$ converges to $-\infty$ as  $n\to\infty$.

On the other hand, in the case $n\in\mathbb{Z}$ with $0\leq n\leq N-1$,
we have $\Delta_n<0$, and the eigenvalues of $\mathcal{A}$ are
\begin{equation}
\begin{gathered}
\lambda_{3n}=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{-\Delta_n}i],\\
\bar{\lambda}_{3n}=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{-\Delta_n}i].
\end{gathered} \label{e2.14}
\end{equation}
For $n\in\mathbb{Z}$ with $ 0\leq n\leq N-1$, we can easily obtain
 $$
[\mathcal{R}e(\lambda_n)+\frac{1}{d}]^2+[\mathcal{I}m(\lambda_n)]^2
=\frac{1}{d^2}.
$$
Moreover, $\Delta_N=0$, and in this case, $\lambda_N=-2/d$.
(the algebraic multiplicity of $\lambda_N$ is two.)
This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.2} \rm
If there exists a positive integer $N$ such that
$$
(N+\frac{1}{2})^2\pi^2<\frac{4}{d^2}-c<(N+\frac{3}{2})^2\pi^2,
$$
in this case, the spectrum $\sigma_p(\mathcal{A})$ has three families:
\begin{align*}
\sigma_p(\mathcal{A})
&=\big\{\lambda_{1n},\; n\geq N+1,\; n\in\mathbb{Z}\big\}
 \cup\big\{\lambda_{2n},\; n\geq N+1,\; n\in\mathbb{Z}\big\} \\
&\quad \cup\big\{\lambda_{3n},\; \bar{\lambda}_{3n},\;
0\leq n\leq N,\; n\in\mathbb{Z}\big\},
\end{align*}
where
\begin{equation}
\begin{gathered}
\lambda_{1n}=\frac{1}{2}[d(\sigma_n^2-c)+\sqrt{\Delta_n}],\quad
 n\in\mathbb{Z},\; n\geq N+1,\\
\lambda_{2n}=\frac{1}{2}[d(\sigma_n^2-c)-\sqrt{\Delta_n}], \quad
 n\in\mathbb{Z},\; n\geq N+1,\\
\lambda_{3n}= \frac{1}{2}[d(\sigma_n^2-c)+\sqrt{-\Delta_n}i],\quad
 n\in\mathbb{Z},\; 0\leq n\leq N.
\end{gathered}\label{e2.15}
\end{equation}
where $\sigma_n$, $\Delta_n$ were given by \eqref{e2.12}.
\end{remark}


\begin{lemma} \label{lem2.2}
Let $\mathcal{A}$ be defined by\eqref{e1.6}. Then
the residual spectrum, $\sigma_r(\mathcal{A})=\emptyset$.
\end{lemma}

\begin{proof}
It is sufficient to show that $\sigma_p(\mathcal{A})=\sigma_p(\mathcal{A}^*)$.
Let us now consider the eigenvalue problem of $\mathcal{A}^*$.
It is seen that $\mathcal{A}^*(f,g)=\lambda(f,g)$, where
$(f,g)\in\mathcal{D}(\mathcal{A}^*)$, $\lambda\in\mathbb{C}$, if
and only if $g(x)=-\lambda f(x)$, and $f(x)\neq0$ satisfies the
 eigenvalue problem
\begin{equation}
\begin{gathered}
(1+d\lambda)f''(x)- (\lambda^2+cd\lambda+c)f(x)=0,\\
f(1)=0,\quad f'(0)=0.
\end{gathered}\label{e2.16}
\end{equation}
It is seen that \eqref{e2.16} is the same with \eqref{e2.4}. Hence,
$\lambda \in\sigma_p(\mathcal{A}^*)$ if and only if
$\lambda \in\sigma_p(\mathcal{A})$. Since the eigenvalues of $\mathcal{A}^*$ are
symmetric with real axis, we have $\sigma_r(\mathcal{A})=\emptyset$.
\end{proof}

Now we are in a position to prove the system operator $\mathcal{A}$
has the Riesz basis property which will lead to
the establishment of the spectrum determined
growth condition, and the exponential stability of the system.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 We first prove that $\{\Phi_n,\; n=0,1,2,\ldots,\}$
is maximal in $\mathcal{H}$, where $\Phi_n$ satisfies
\begin{gather*}
\Phi_{1n}=\sqrt{\frac{2\lambda_{1n}}{d(\sigma_n^2-c)}}
(\frac{1}{\lambda_{1n}}\cos[(n+\frac{1}{2})\pi x],
\cos[(n+\frac{1}{2})\pi x]),
\quad n\in\mathbb{Z},\ n\geq N+1,\\
\Phi_{2n}= \sqrt{\frac{2\lambda_{2n}}{d(\sigma_n^2-c)}}
(\frac{1}{\lambda_{2n}}\cos[(n+\frac{1}{2})\pi x],\cos[(n+\frac{1}{2})\pi x]),
\quad n\in\mathbb{Z},\; n\geq N+1,\\
\Phi_{3n}=(\frac{1}{\lambda_{3n}}\cos[(n+\frac{1}{2})\pi x],
 \cos[(n+\frac{1}{2})\pi x]),
\quad n\in\mathbb{Z},\; 0\leq n\leq N-1,\\
\bar{\Phi}_{3n}= (\frac{1}{\bar{\lambda}_{3n}}\cos[(n+\frac{1}{2})\pi x],
 \cos[(n+\frac{1}{2})\pi x]),
\quad n\in\mathbb{Z},\; 0\leq n\leq N-1,\\
\Phi_N=(\frac{1}{\lambda_{N}}\cos[(N+\frac{1}{2})\pi x],\cos[(N+\frac{1}{2})\pi x]),\\
\tilde{\Phi}_N=\frac{1}{\sqrt{\frac{4}{d^2}+
\frac{d^2}{8}-1}}(\cos[(N+\frac{1}{2})\pi x],
 (\frac{d}{2}-\frac{2}{d})\cos[(N+\frac{1}{2})\pi x]),
\end{gather*}
and $\tilde{\Phi}_N$ is the generalized eigenvector of $\lambda_N$.

Suppose that $z=(z_1,z_2)\in\mathcal{H}$ is orthogonal to every $\Phi_n$.
Then
$$
\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\frac{1}{\lambda_{n}}
 \cos[(n+\frac{1}{2})\pi x)]}
 {\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle=0,\quad \forall n\in\mathbb{Z},\; n\geq0,\; n\neq N,
$$
i.e.,
\begin{equation}
\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(n+\frac{1}{2})\pi x)]}
 {\lambda_{n}\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle=0,\quad \forall n\in\mathbb{Z},\; n\geq0,\ ;n\neq N.\label{e2.17}
\end{equation}
And
\begin{gather}
\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\frac{1}{\lambda_{N}}
\cos[(N+\frac{1}{2})\pi x)]}
 {\cos[(N+\frac{1}{2})\pi x)]}
\Big\rangle=0,\label{e2.18} \\
\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(N+\frac{1}{2})\pi x]}
{(\frac{d}{2}-\frac{2}{d})\cos[(N+\frac{1}{2})\pi x]}\Big\rangle=0,\label{e2.19}
\end{gather}
where  $\lambda_N=-2/d$.
Next we distinguish three cases.
\smallskip

\noindent\textbf{Case 1.}
 For $n\in\mathbb{Z},\ n\geq N+1$, noticing that $\lambda_{1n}<0$, $\lambda_{2n}<0$,
\begin{align*}
&\int_0^1(\lambda_{1n}-\lambda_{2n})z_2(x)\cos[(n+\frac{1}{2})\pi x]
 dx\\
&=\Big \langle\binom{z_1(x)}{z_2(x)},\binom{0}
 {(\lambda_{1n}-\lambda_{2n})\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle\\
&=\sqrt{\frac{d(\sigma_n^2-c)\lambda_{2n}}{2}}\langle z,\Phi_{2n}\rangle
 -\sqrt{\frac{d(\sigma_n^2-c)\lambda_{1n}}{2}}\langle z,\Phi_{1n}\rangle
=0,
\end{align*}
where $\lambda_{1n}-\lambda_{2n}=\sqrt{\Delta_n}>0$.
Then for any $n\in\mathbb{Z}$ with $n\geq N+1$,
\begin{equation}
\int_0^1z_2(x)\cos[(n+\frac{1}{2})\pi x]  dx =0.\label{e2.20}
\end{equation}
From\eqref{e2.17}, \eqref{e2.20} and for any $n\in\mathbb{Z}$ with $n\geq N+1$,
\begin{align*}
0&=\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(n+\frac{1}{2})\pi x)]}
 {\lambda_{1n}\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle\\
&=\int_0^1-(n+\frac{1}{2})\pi z'_1(x)\sin[(n+\frac{1}{2})\pi x)]dx\\
&\quad +c\int_0^1z_1(x)\cos[(n+\frac{1}{2})\pi x)]dx\\
&\quad +\int_0^1\lambda_{1n}z_2(x)\cos[(n+\frac{1}{2})\pi x)]dx\\
&=(c-\sigma_n^2)\int_0^1z_1(x)\cos[(n+\frac{1}{2})\pi x)]dx,
\end{align*}
i.e.,
\begin{equation}
\int_0^1z_1(x)\cos[(n+\frac{1}{2})\pi x)]dx=0,\label{e2.21}
\end{equation}
for any $n\in\mathbb{Z}$, $n\geq N+1$.
\smallskip

\noindent\textbf{Case 2.}
For $0\leq n\leq N-1$, $n\in\mathbb{Z}$,
\begin{align*}
&(\lambda_{3n}-\bar{\lambda}_{3n})
\int_0^1z_2(x)\cos[(n+\frac{1}{2})\pi x)]dx\\
& =\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(n+\frac{1}{2})\pi x)]}
 {\lambda_{3n}\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle
 -\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(n+\frac{1}{2})\pi x)]}
 {\bar{\lambda}_{3n}\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle\\
&=\lambda_{3n}\langle z,\Phi_{3n}\rangle-\bar{\lambda}_{3n}\langle z,\bar{\Phi}_{3n}
 \rangle= 0,
\end{align*}
where $\lambda_{3n}-\bar{\lambda}_{3n}=\sqrt{-\Delta_n}i$.
Then for any $0\leq n\leq N-1$ with $n\in\mathbb{Z}$,
\begin{equation}
\int_0^1z_2(x)\cos[(n+\frac{1}{2})\pi x)]dx=0.\label{e2.22}
\end{equation}
Similarly,
\begin{equation}
\begin{aligned}
0&=\Big\langle\binom{z_1(x)}{z_2(x)},\binom{\cos[(n+\frac{1}{2})\pi x)]}
 {\lambda_{3n}\cos[(n+\frac{1}{2})\pi x)]}
\Big\rangle\\
&=(c-\sigma_n^2)\int_0^1z_1(x)\cos[(n+\frac{1}{2})\pi x)]dx \\
&\quad +\bar{\lambda}_{3n}\int_0^1z_2(x)\cos[(n+\frac{1}{2})\pi x)]dx.
 \end{aligned}\label{e2.23}
\end{equation}
Combining \eqref{e2.22} and\eqref{e2.23},  for $0\leq n\leq N-1,\ n\in\mathbb{Z}$,
we have
\begin{equation}
\int_0^1z_1(x)\cos[(n+\frac{1}{2})\pi x)]dx=0.\label{e2.24}
\end{equation}
\smallskip

\noindent\textbf{Case 3.}
Multiplying \eqref{e2.19} by $d/2$ and adding it to \eqref{e2.18}, we obtain
\[
0=\Big\langle\binom{z_1(x)}{z_2(x)},\binom{0}
 {\frac{4}{d^2}\cos[(N+\frac{1}{2})\pi x)]}
\Big\rangle
=\frac{d^2}{4}\int_0^1z_2(x)\cos[(N+\frac{1}{2})\pi x)]dx,
\]
i.e.,
\begin{equation}
\int_0^1z_2(x)\cos[(N+\frac{1}{2})\pi x)]dx=0.\label{e2.25}
\end{equation}
Combining \eqref{e2.19} and\eqref{e2.25}, we can easily obtain
\begin{equation}
\int_0^1z_1(x)\cos[(N+\frac{1}{2})\pi x)]dx=0,\label{e2.26}
\end{equation}
Since $\{\cos[(n+\frac{1}{2})\pi x)],\ n\in\mathbb{Z},\ n\geq0\}$ is
maximal in $L^2(0,1)$, and considering \eqref{e2.20}-\eqref{e2.22} and
\eqref{e2.24}-\eqref{e2.26}, we have that $z=(z_1,z_2)=0$.
 Thus $\{\Phi_n,\; n\in\mathbb{Z},\; n\geq0\}$ is maximal in $\mathcal{H}$.

For any $n, k\geq N+1$, $n\in\mathbb{Z}$,
\begin{align*}
&\Big\langle\binom{\frac{1}{\lambda_{1n}}\cos[(n+\frac{1}{2})\pi x)]}
 {\cos[(n+\frac{1}{2})\pi x)]},\binom{\frac{1}{\lambda_{1k}}\cos[(k+\frac{1}{2})\pi x)]}
 {\cos[(k+\frac{1}{2})\pi x)]}\Big\rangle\\
&=\frac{(n+\frac{1}{2})(k+\frac{1}{2})\pi^2}{\lambda_{1n}\lambda_{1k}}\int_0^1
 \sin[(n+\frac{1}{2})\pi x)]\sin[(k+\frac{1}{2})\pi x)]dx\\
&\quad +\frac{c}{\lambda_{1n}\lambda_{1k}}\int_0^1
 \cos[(n+\frac{1}{2})\pi x)]\cos[(k+\frac{1}{2})\pi x)]dx\\
&\quad +\int_0^1  \cos[(n+\frac{1}{2})\pi x)]\cos[(k+\frac{1}{2})\pi x)]dx \\
&= \frac{(n+\frac{1}{2})(k+\frac{1}{2})\pi^2}{2\lambda_{1n}\lambda_{1k}}\int_0^1
 \cos[(n-k)\pi x]-\cos[(n+k+1)\pi x]dx\\
&\quad +\frac{1}{2}(\frac{c}{\lambda_{1n}\lambda_{1k}}+1)
 \int_0^1  \cos[(n-k)\pi x]+\cos[(n+k+1)\pi x]dx\\
&= \begin{cases}
 0,&n\neq k,\\
\frac{c-\sigma_n^2+\lambda^2_{1n}}
 {2\lambda^2_{1n}}, &n=k.
\end{cases}\\
&=\begin{cases}
 0, &n\neq k,\\
\frac{d(\sigma_n^2-c)} {2\lambda_{1n}}, &n=k.
\end{cases}
 \end{align*}
Then
\begin{equation}
\begin{gathered}
\langle\Phi_{1n},\Phi_{1k}\rangle=0,\quad
  n,\; k\geq N+1,\; n\in\mathbb{Z},\; n\neq k,\\
\langle\Phi_{1n},\Phi_{1n}\rangle=1, \quad n\geq N+1,\; n\in\mathbb{Z}.
\end{gathered}   \label{e2.27}
\end{equation}

Similarly for any $n$, $k\geq N+1$, $n\in\mathbb{Z}$, we have
\begin{align*}
&\Big\langle\binom{\frac{1}{\lambda_{1n}}\cos[(n+\frac{1}{2})\pi x)]}
 {\cos[(n+\frac{1}{2})\pi x)]},\binom{\frac{1}{\lambda_{2k}}
 \cos[(k+\frac{1}{2})\pi x)]}
 {\cos[(k+\frac{1}{2})\pi x)]}\Big\rangle\\
&=\frac{(n+\frac{1}{2})(k+\frac{1}{2})\pi^2}{\lambda_{1n}\lambda_{2k}}\int_0^1
 \sin[(n+\frac{1}{2})\pi x)]\sin[(k+\frac{1}{2})\pi x)]dx\\
&\quad +\frac{c}{\lambda_{1n}\lambda_{2k}}\int_0^1
 \cos[(n+\frac{1}{2})\pi x)]\cos[(k+\frac{1}{2})\pi x)]dx\\
&\quad +\int_0^1 \cos[(n+\frac{1}{2})\pi x)]\cos[(k+\frac{1}{2})\pi x)]dx \\
&= \frac{(n+\frac{1}{2})(k+\frac{1}{2})\pi^2}{2\lambda_{1n}\lambda_{2k}}\int_0^1
 \cos[(n-k)\pi x]-\cos[(n+k+1)\pi x]dx\\
&\quad +\frac{1}{2}(\frac{c}{\lambda_{1n}\lambda_{2k}}+1)
 \int_0^1  \cos[(n-k)\pi x]+\cos[(n+k+1)\pi x]dx.
 \end{align*}
Then
\begin{equation}
\langle\Phi_{1n},\Phi_{2k}\rangle=0,\quad n\neq k,
\; n,\ ;k\geq N+1,\; n\in\mathbb{Z},\label{e2.28}
\end{equation}
 and by the fact that $\lambda_{1n}\lambda_{2n}=c-\sigma_n^2$,
 where $\sigma_n=(n+\frac{1}{2})\pi i$, we have
\begin{equation}
\langle\Phi_{1n},\Phi_{2n}\rangle
=\sqrt{\frac{2\lambda_{1n}}{d(\sigma_n^2-c)}}
 \sqrt{\frac{2\lambda_{2n}}{d(\sigma_n^2-c)}}
 \frac{c-\sigma_n^2+\lambda_{1n}\lambda_{2n}}{2\lambda_{1n}\lambda_{2n}}
 =\frac{2}{d\sqrt{c-\sigma_n^2}}.\label{e2.29}
\end{equation}
For any $n\geq N+1$, $0\leq m\leq N-1$, $n, m\in\mathbb{Z}$,
\begin{align*}
&\Big\langle\binom{\frac{1}{\lambda_{1n}}\cos[(n+\frac{1}{2})\pi x)]}
 {\cos[(n+\frac{1}{2})\pi x)]},\binom{\frac{1}{\lambda_{3m}}
 \cos[(m+\frac{1}{2})\pi x)]}{\cos[(m+\frac{1}{2})\pi x)]}\Big\rangle\\
&=\frac{(n+\frac{1}{2})(m+\frac{1}{2})\pi^2}{\lambda_{1n}\bar{\lambda}_{3m}}\int_0^1
 \sin[(n+\frac{1}{2})\pi x)]\sin[(m+\frac{1}{2})\pi x)]dx\\
&\quad +\frac{c}{\lambda_{1n}\bar{\lambda}_{3m}}\int_0^1
 \cos[(n+\frac{1}{2})\pi x)]\cos[(m+\frac{1}{2})\pi x)]dx\\
&\quad +\int_0^1  \cos[(n+\frac{1}{2})\pi x)]\cos[(m+\frac{1}{2})\pi x)]dx \\
&= \frac{(n+\frac{1}{2})(m+\frac{1}{2})\pi^2}{2\lambda_{1n}\bar{\lambda}_{3m}}
 \int_0^1 \cos[(n-m)\pi x]-\cos[(n+m+1)\pi x]dx\\
&\quad +\frac{1}{2}(\frac{c}{\lambda_{1n}\bar{\lambda}_{3m}}+1)
 \int_0^1
 \cos[(n-m)\pi x]+\cos[(n+m+1)\pi x]dx
=0,
 \end{align*}
i.e.,
\begin{equation}
\langle\Phi_{1n},\Phi_{3m}\rangle=0.\label{e2.30}
\end{equation}
Furthermore,
\begin{equation}
\begin{aligned}
\langle\Phi_{3m},\bar{\Phi}_{3m}\rangle
&=\Big\langle\binom{\frac{1}{\lambda_{3m}}\cos[(n+\frac{1}{2})\pi x)]}
 {\cos[(n+\frac{1}{2})\pi x)]},\binom{\frac{1}{{\bar{\lambda}}_{3m}}
 \cos[(m+\frac{1}{2})\pi x)]}{\cos[(m+\frac{1}{2})\pi x)]}\Big\rangle\\
&=\frac{(m+\frac{1}{2})^2\pi^2}{(\lambda_{3m})^2}\int_0^1
 \sin^2[(m+\frac{1}{2})\pi x)]dx\\
&\quad +(\frac{c}{(\lambda_{3m})^2}+1)\int_0^1
 \cos^2[(m+\frac{1}{2})\pi x)]dx\\
&= \frac{(m+\frac{1}{2})^2\pi^2}{2(\lambda_{3m})^2}
 +\frac{1}{2}(\frac{c}{(\lambda_{3m})^2}+1)\\
&= \frac{(m+\frac{1}{2})^2\pi^2+c+(\lambda_{3m})^2}{2(\lambda_{3m})^2},
 \end{aligned}\label{e2.31}
\end{equation}
 where $\lambda_{3m}=\frac{1}{2}[d(\sigma_m^2-c)+\sqrt{-\Delta_m}i]$,
$0\leq m\leq N-1$, $m\in\mathbb{Z}$, and
\begin{gather*}
 \sigma_m=i(m+\frac{1}{2})\pi,\\
\Delta_m  =d^2(c-\sigma_m^2)^2-4(c-\sigma_m^2).
\end{gather*}
Moreover, $|\lambda_{3m}|^2=c-\sigma_m^2$. Then
\begin{gather}
\langle\Phi_N,\Phi_N\rangle
=\Big\langle\binom{\frac{1}{\lambda_{N}}\cos[(N+\frac{1}{2})\pi x]}
{\cos[(N+\frac{1}{2})\pi x]},\binom{\frac{1}{\lambda_{N}}
\cos[(N+\frac{1}{2})\pi x]}{\cos[(N+\frac{1}{2})\pi x]}\Big\rangle=1.
\label{e2.32}\\
\begin{aligned}
\langle\tilde{\Phi}_N,\tilde{\Phi}_N\rangle
&=\frac{1}{\frac{4}{d^2}+ \frac{d^2}{8}-1}
\Big\langle\binom{\cos[(N+\frac{1}{2})\pi x]}{(\frac{d}{2}-\frac{2}{d})
 \cos[(N+\frac{1}{2})\pi x]}, \\
&\quad \binom{\cos[(N+\frac{1}{2})\pi x]}{(\frac{d}{2}-\frac{2}{d})
 \cos[(N+\frac{1}{2})\pi x]}\Big\rangle=1.
\end{aligned}\label{e2.33}\\
\begin{aligned}
\langle\Phi_N,\tilde{\Phi}_N\rangle
&=\frac{1}{\sqrt{\frac{4}{d^2}+\frac{d^2}{8}-1}}
\Big\langle\binom{\frac{1}{\lambda_{N}}\cos[(N+\frac{1}{2})\pi x]}
 {\cos[(N+\frac{1}{2})\pi x]},
 \binom{\cos[(N+\frac{1}{2})\pi x]}{(\frac{d}{2}-\frac{2}{d})
 \cos[(N+\frac{1}{2})\pi x]}\Big\rangle\\
&=\frac{d^2-8}{\sqrt{2d^4-16d^2+64}}.
\end{aligned}\label{e2.34}
\end{gather}
Similarly,  for any $n$, $k\geq N+1$, $0\leq m$, $l\leq N-1$,
$n, k, m, l\in\mathbb{Z}$, the following holds
\begin{gather*}
\langle\Phi_{1n},\bar{\Phi}_{3m}\rangle=0,\quad
 \langle\Phi_{1n},\Phi_N\rangle=0,\quad
\langle\Phi_{1n},\tilde{\Phi}_N\rangle=0,\\
\langle\Phi_{2n},\Phi_{2k}\rangle=0\;(n\neq k),\quad
\langle\Phi_{2n},\Phi_{2n}\rangle=1,\quad
 \langle\Phi_{2n},\Phi_{3m}\rangle=0,\\
 \langle\Phi_{2n},\bar{\Phi}_{3m}\rangle=0,\quad
  \langle\Phi_{2n},\Phi_N\rangle=0,\quad
 \langle\Phi_{2n},\tilde{\Phi}_N\rangle=0,\\
 \langle\Phi_{3m},\Phi_{3l}\rangle=0\; (m\neq l),\quad
 \langle\Phi_{3m},\Phi_{3m}\rangle=\frac{c-\sigma_m^2+|\lambda_{3m}|^2}
 {2|\lambda_{3m}|^2}=1,\\
 \langle\Phi_{3m},\bar{\Phi}_{3m}\rangle=\frac{c-\sigma_m^2+\lambda^2_{3m}}
 {2\lambda^2_{3m}},\quad
 \langle\Phi_{3m},\bar{\Phi}_{3l}\rangle=0\; ( m\neq l),\\
  \langle\Phi_{3m},\Phi_N\rangle=0,\quad
 \langle{\Phi}_{3m},\tilde{\Phi}_N\rangle=0,\quad
 \langle\bar{\Phi}_{3m},\bar{\Phi}_{3l}\rangle=0\; (m\neq l),\\
 \langle\bar{\Phi}_{3m},\bar{\Phi}_{3m}\rangle=
 \frac{c-\sigma_m^2+|\bar{\lambda}_{3m}|^2}
 {2|\bar{\lambda}_{3m}|^2}=1,\quad
 \langle\bar{\Phi}_{3m},\Phi_N\rangle=0,\quad
 \langle\bar{\Phi}_{3m},\tilde{\Phi}_N\rangle=0.\\
 \end{gather*} %\label{e2.35}
From the above equalities we  deduce that
\begin{equation}
\|\sum_{m=0}^{N-1}\alpha_m\Phi_{3m}\|^2=
\sum_{m=0}^{N-1}|\alpha_m|^2,\quad
\|\sum_{m=0}^{N-1}\beta_m\bar{\Phi}_{3m}\|^2=
\sum_{m=0}^{N-1}|\beta_m|^2. \label{e2.36}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\|\sum_{m=0}^{N-1}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2\\
&=\Big\langle\sum_{m=0}^{N-1}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m},
\sum_{l=0}^{N-1}\alpha_l\Phi_{3l}+\beta_l\bar{\Phi}_{3l}\Big\rangle\\
&=\sum_{m=0}^{N-1}(|\alpha_m|^2\|\Phi_{3m}\|^2+|\beta_m|^2
 \|\bar{\Phi}_{3m}\|^2) \\
&\quad +\sum_{m=0}^{N-1}\alpha_m\bar{\beta}_m
\langle\Phi_{3m},\bar{\Phi}_{3m}\rangle
+\beta_m\bar{\alpha}_m\langle\bar{\Phi}_{3m},{\Phi}_{3m}\rangle\\
&=\sum_{m=0}^{N-1}(|\alpha_m|^2+|\beta_m|^2)+\sum_{m=0}^{N-1}
2\mathcal{R}e(\alpha_m\bar{\beta}_m\langle{\Phi}_{3m},\bar{\Phi}_{3m}\rangle).
\end{aligned}\label{e2.37}
\end{equation}
A direct calculation from \eqref{e2.31} indicates, for any
$0\leq m\leq N-1$ and $m\in\mathbb{Z}$, we have
\begin{equation}
|\langle{\Phi}_{3m},\bar{\Phi}_{3m}\rangle|
=\frac{d}{2}\sqrt{c-\sigma_m^2},\label{e2.38}
\end{equation}
and hence
\begin{equation}
|2\mathcal{R}e
(\alpha_m\bar{\beta_m}\langle{\Phi}_{3m},\bar{\Phi}_{3m}\rangle|\leq
(|\alpha_m|^2+|\bar{\beta}_m|^2)\frac{d}{2}\sqrt{c-\sigma_m^2}.\label{e2.39}
\end{equation}
Combining \eqref{e2.37} and \eqref{e2.39}, we obtain
\begin{equation}
\|\sum_{m=0}^{N-1}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2
\leq(1+\frac{d}{2}\sqrt{c-\sigma_{N-1}^2})
\sum_{m=0}^{N-1}(|\alpha_m|^2+|\beta_m|^2),\label{e2.40}
\end{equation}
and
\begin{equation}
\begin{aligned}
\|\sum_{m=0}^{N-1}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2
&\geq\sum_{m=0}^{N-1}(1-\frac{d}{2}
 \sqrt{c-\sigma_{m}^2})(|\alpha_m|^2+|\beta_m|^2)\\
&>(1-\frac{d}{2}\sqrt{c-\sigma_{N-1}^2})
\sum_{m=0}^{N-1}(|\alpha_m|^2+|\beta_m|^2),
\end{aligned}\label{e2.41}
\end{equation}
where $1-\frac{d}{2}\sqrt{c-\sigma_{N-1}^2}>1-\frac{d}{2}\sqrt{c-\sigma_{N}^2}=0$.

Without loss of generality, let $K_1, K_2\in\mathbb{Z}$,
$K_1\geq K_2>0$,
\begin{equation}
\begin{aligned}
&\|\sum_{n=N+1}^{N+K_1}\alpha_n\Phi_{1n}+\sum_{n=N+1}^{N+K_2}
\beta_n{\Phi}_{2n}\|^2\\
&=\sum_{n=N+1}^{N+K_1}|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2\\
&\quad +\sum_{n=N+1}^{N+K_2}\langle\alpha_n\Phi_{1n},\beta_n{\Phi}_{2n}\rangle+
\langle\beta_n{\Phi}_{2n},\alpha_n\Phi_{1n}\rangle.
\end{aligned}\label{e2.42}
\end{equation}
While
\begin{equation}
\begin{aligned}
\langle\alpha_n\Phi_{1n},\beta_n{\Phi}_{2n}\rangle+
\langle\beta_n{\Phi}_{2n},\alpha_n\Phi_{1n}\rangle
&=2\mathcal{R}e(\alpha_n\beta_n\langle\Phi_{1n},{\Phi}_{2n}\rangle\\
&\leq2|\alpha_n||\beta_n||\langle\Phi_{1n},{\Phi}_{2n}\rangle|\\
&\leq\frac{2}{d\sqrt{c-\sigma_n^2}}(|\alpha_n|^2+|\beta_n|^2)\\
&\leq\frac{2}{d\sqrt{c-\sigma_{N+1}^2}}
(|\alpha_n|^2+|\beta_n|^2).
\end{aligned}\label{e2.43}
\end{equation}
Inserting \eqref{e2.43} into \eqref{e2.42}, we obtain
\begin{equation}
\begin{aligned}
&\|\sum_{n=N+1}^{N+K_1}\alpha_n\Phi_{1n}+\sum_{n=N+1}^{N+K_2}
\beta_n{\Phi}_{2n}\|^2\\
&\leq \sum_{n=N+1}^{N+K_1}|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2
+\sum_{n=N+1}^{N+K_2}\frac{2}{d\sqrt{c-\sigma_n^2}}(|\alpha_n|^2+|\beta_n|^2)
\\
&<\sum_{n=N+1}^{N+K_1}(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})|\alpha_n|^2
+\sum_{n=N+1}^{N+K_2}(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})|\beta_n|^2
\\
&=(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})[\sum_{n=N+1}^{N+K_1}
|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2],
\end{aligned}\label{e2.44}
\end{equation}
and
\begin{equation}
\begin{aligned}
&\|\sum_{n=N+1}^{N+K_1}\alpha_n\Phi_{1n}+\sum_{n=N+1}^{N+K_2}
\beta_n{\Phi}_{2n}\|^2\\
&\geq \sum_{n=N+1}^{N+K_1}|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2
-\sum_{n=N+1}^{N+K_2}\frac{2}{d\sqrt{c-\sigma_n^2}}(|\alpha_n|^2+|\beta_n|^2)
\\
&>(1-\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})[\sum_{n=N+1}^{N+K_1}|\alpha_n|^2
+\sum_{n=N+1}^{N+K_2}|\beta_n|^2],
\end{aligned}\label{e2.45}
\end{equation}
where $1-\frac{2}{d\sqrt{c-\sigma_{N+1}^2}}>1-\frac{2}{d\sqrt{c-\sigma_{N}^2}}=0$.
Similarly,
\begin{equation}
\begin{aligned}
&\|\alpha_N\Phi_N+\tilde{\alpha}_N\tilde{\Phi}_N\|^2 \\
&=|\alpha_N|^2\|\Phi_N\|^2+|\tilde{\alpha}_N|^2\|\tilde{\Phi}_N\|^2
+2\mathcal{R}e({\alpha}_N\tilde{\alpha}_N\langle\Phi_N,\tilde{\Phi}_N\rangle),
\end{aligned}\label{e2.46}
\end{equation}
and
\begin{equation}
\begin{aligned}
|2\mathcal{R}e({\alpha}_N\tilde{\alpha}_N\langle\Phi_N,\tilde{\Phi}_N\rangle)|
&\leq(|\alpha_N|^2+|\tilde{\alpha}_N|^2)|\langle\Phi_N,\tilde{\Phi}_N\rangle|\\
&=\frac{|d^2-8|}{\sqrt{2d^4-16d^2+64}}(|\alpha_N|^2+|\tilde{\alpha}_N|^2|).
\end{aligned}\label{e2.47}
\end{equation}
Combining \eqref{e2.46} and \eqref{e2.47},
we  obtain 
\begin{gather}
\|\alpha_N\Phi_N+\tilde{\alpha}_N\tilde{\Phi}_N\|^2\leq
(1+\frac{|d^2-8|}{\sqrt{2d^4-16d^2+64}})(|\alpha_N|^2+|\tilde{\alpha}_N|^2|),
\label{e2.48}\\
\|\alpha_N\Phi_N+\tilde{\alpha}_N\tilde{\Phi}_N\|^2\geq
(1-\frac{|d^2-8|}{\sqrt{2d^4-16d^2+64}})(|\alpha_N|^2+|\tilde{\alpha}_N|^2|),
\label{e2.49}
\end{gather}
where 
\[
1-\frac{|d^2-8|}{\sqrt{2d^4-16d^2+64}}>0.
\]
Considering\eqref{e2.40}, \eqref{e2.41}, \eqref{e2.44}, \eqref{e2.45} 
\eqref{e2.48}, and \eqref{e2.49},
by Definition \ref{def1.1} we  conclude that
$$
\big\{\Phi_{1n}, \Phi_{2n}, \Phi_{3m}, \bar{\Phi}_{3m},
\Phi_N, \tilde{\Phi}_N,\; n\geq N+1,\, 0\leq m\leq N-1,\, n, m\in\mathbb{Z}\big\}
$$
form a Riesz basis. Furthermore  the spectrum-determined growth
condition holds for the semigroup $e^{\mathcal{A}t}$ generated
by $\mathcal{A}$. This completes the proof of Theorem \ref{thm1.2}.
Hence semigroup $e^{\mathcal{A}t}$ generated
by $\mathcal{A}$ is exponentially stable for $d^2c<4$.
\end{proof}

\begin{remark} \label{rmk2.4} \rm
If there exists a positive integer $N$ such that
\begin{equation}
(N+\frac{1}{2})^2\pi^2<\frac{4}{d^2}-c<(N+\frac{3}{2})^2\pi^2,\label{e2.50}
\end{equation}
the operator $\mathcal{A}$ has the eigenvectors
\[
\big\{\Phi_{1n}, \Phi_{2n}, \Phi_{3m}, \bar{\Phi}_{3m},\; n\geq N+1,\,
0\leq m\leq N,\, n, m\in\mathbb{Z}\big\},
\]
where
\begin{equation}
\begin{gathered}
\Phi_{1n}=\sqrt{\frac{2\lambda_{1n}}{d(\sigma_n^2-c)}}
(\frac{1}{\lambda_{1n}}\cos[(n+\frac{1}{2})\pi x],
\cos[(n+\frac{1}{2})\pi x]),
 \quad n\in\mathbb{Z},\; n\geq N+1,\\
\Phi_{2n}=\sqrt{\frac{2\lambda_{2n}}{d(\sigma_n^2-c)}}
(\frac{1}{\lambda_{2n}}\cos[(n+\frac{1}{2})\pi x],\cos[(n+\frac{1}{2})\pi x]),
\quad n\in\mathbb{Z},\ n\geq N+1,\\
\Phi_{3m}=(\frac{1}{\lambda_{3m}}\cos[(m+\frac{1}{2})\pi x],
\cos[(m+\frac{1}{2})\pi x]),\quad m\in\mathbb{Z},\; 0\leq m\leq N,\\
\bar{\Phi}_{3m}=(\frac{1}{\bar{\lambda}_{3m}}\cos[(m+\frac{1}{2})\pi x],
\cos[(m+\frac{1}{2})\pi x]),\quad n\in\mathbb{Z},\; 0\leq n\leq N,
\end{gathered}\label{e2.51}
\end{equation}
and $\lambda_n$ were given by\eqref{e2.15}.
\end{remark}

First we can easily check that
$$
{\overline{\rm span}}\{\Phi_{1n}, \Phi_{2n}, \Phi_{3m}, \bar{\Phi}_{3m},\; n\geq N+1,\,
0\leq m\leq N,\ n,\ m\in\mathbb{Z}\}=\mathcal{H}.
$$
Furthermore, for any $n, k\geq N+1$, $0\leq m$, $l\leq N$, $n, k, m, l\in\mathbb{Z}$, 
the following holds:
\begin{gather*}
\langle\Phi_{1n},\Phi_{1k}\rangle=0\;(n\neq k),\quad
  \langle\Phi_{1n},\Phi_{1n}\rangle=1,\quad
 \langle\Phi_{1n},\Phi_{2n}\rangle=\frac{2}{d\sqrt{c-\sigma_n^2}},\\
 \langle\Phi_{1n},\Phi_{2k}\rangle=0\; (n\neq k),\quad
\langle\Phi_{2n},\Phi_{2n}\rangle=1,\quad
 \langle\Phi_{2 n},\Phi_{2k}\rangle=0\;(n\neq k),\\
 \langle\Phi_{1n},{\Phi}_{3m}\rangle=0,\quad
 \langle\Phi_{1n},\bar{\Phi}_{3m}\rangle=0,\quad
 \langle\Phi_{2n},\Phi_{3m}\rangle=0, \\
 \langle\Phi_{2n},\bar{\Phi}_{3m}\rangle=0,\quad
 \langle\Phi_{3m},\Phi_{3m}\rangle=1,\quad
 \langle\Phi_{3m},\Phi_{3l}\rangle=0\;(m\neq l),\\
 \langle\bar{\Phi}_{3m},\bar{\Phi}_{3m}\rangle=1,\quad
 \langle\bar{\Phi}_{3m},\bar{\Phi}_{3l}\rangle=0\; (m\neq l),\quad
 \langle\Phi_{3m},\bar{\Phi}_{3l}\rangle=0\;(m\neq l),\\
 \langle\Phi_{3m},\bar{\Phi}_{3m}\rangle=
 \frac{c-\sigma_m^2+\lambda^2_{3m}} {2\lambda^2_{3m}}.
 \end{gather*}
 Let $K_1, K_2\in\mathbb{Z}$, $K_1\geq K_2>0$. Then
\begin{equation}
\begin{aligned}
&\|\sum_{n=N+1}^{N+K_1}\alpha_n\Phi_{1n}+\sum_{n=N+1}^{N+K_2}\beta_n{\Phi}_{2n}\|^2\\
&\leq \sum_{n=N+1}^{N+K_1}|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2
+\sum_{n=N+1}^{N+K_2}\frac{2}{d\sqrt{c-\sigma_n^2}}(|\alpha_n|^2+|\beta_n|^2)
\\
&<\sum_{n=N+1}^{N+K_1}(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})|\alpha_n|^2
+\sum_{n=N+1}^{N+K_2}(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})|\beta_n|^2
\\
&=(1+\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})[\sum_{n=N+1}^{N+K_1}
|\alpha_n|^2+\sum_{n=N+1}^{N+K_2}|\beta_n|^2].
\end{aligned}\label{e2.52}
\end{equation}
Similarly, we can deduce that
\begin{equation}
\begin{aligned}
&\|\sum_{n=N+1}^{N+K_1}\alpha_n\Phi_{1n}+\sum_{n=N+1}^{N+K_2}
\beta_n{\Phi}_{2n}\|^2 \\
&\geq(1-\frac{2}{d\sqrt{c-\sigma_{N+1}^2}})[
\sum_{n=N+1}^{N+K_1}|\alpha_n|^2
+\sum_{n=N+1}^{N+K_2}|\beta_n|^2],
\end{aligned}\label{e2.53}
\end{equation}
where $1-\frac{2}{d\sqrt{c-\sigma_{N+1}^2}}>0$.
Moreover,
\begin{align*}
\|\sum_{m=0}^{N}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2
&=\langle\sum_{m=0}^{N}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m},
\sum_{l=0}^{N}\alpha_l\Phi_{3l}+\beta_l\bar{\Phi}_{3l}\rangle\\
&=\sum_{m=0}^{N}(|\alpha_m|^2\|\Phi_{3m}\|^2
+|\beta_m|^2\|\bar{\Phi}_{3m}\|^2)\\
&\quad +\sum_{m=0}^{N}\alpha_m\bar{\beta}_m
\langle\Phi_{3m},\bar{\Phi}_{3m}\rangle
+\beta_m\bar{\alpha}_m\langle\bar{\Phi}_{3m},{\Phi}_{3m}\rangle\\
&=\sum_{m=0}^{N}(|\alpha_m|^2+|\beta_m|^2)+\sum_{m=0}^{N}
2\mathcal{R}e(\alpha_m\bar{\beta}_m\langle{\Phi}_{3m},\bar{\Phi}_{3m}\rangle).
\end{align*}
By estimating the term 
$2\mathcal{R}e(\alpha_m\bar{\beta}_m\langle{\Phi}_{3m},\bar{\Phi}_{3m}\rangle)$,
we can obtain 
\begin{gather*}
\|\sum_{m=0}^{N}\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2
\leq(1+\frac{d}{2}\sqrt{c-\sigma_N^2})\sum_{m=0}^{N}(|\alpha_m|^2+|\beta_m|^2),\\
\|\sum_{m=0}^{N}(\alpha_m\Phi_{3m}+\beta_m\bar{\Phi}_{3m}\|^2)\geq
(1-\frac{d}{2}\sqrt{c-\sigma_N^2}){\sum_{m=0}^{N}}
|\alpha_m|^2+|\beta_m|^2,
\end{gather*}
where $1-\frac{d}{2}\sqrt{c-\sigma_N^2}>0$.
Combining the above inequalities with \eqref{e2.52}, and \eqref{e2.53}, 
we conclude that, in the case of \eqref{e2.51},
\[
\{\Phi_{1n}, \Phi_{2n}, \Phi_{3m}, \bar{\Phi}_{3m},\; n\geq N+1,\,
0\leq m\leq N,\, n, m\in\mathbb{Z}\},
\]
also forms a Riesz basis of the state space $\mathcal{H}$.

\subsection*{Acknowledgments}
This work is supported by the National Science
Foundation of China (Nos. 11401351, 11871315)
 and by the Youth Science  Foundation of Shanxi Province(201601D021010).
The authors would like to thank the anonymous referees
and associate editor for their very helpful suggestions and comments.

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